Shear reinforcement design on detailing provisions for column
Minimal shear reinforcement according to detailing provisions for beam is calculated according to formula
Aswm,min = max(Aswc,ϕ,min; Aswc,long,min)
where
Aswc,ϕ,min | minimal area of shear reinforcement from minimal bar diameter 9.5.3(1) |
Aswc,long,min | minimal area of shear reinforcement from maximum longitudinal spacing 9.5.3(3) |
Minimal area of shear reinforcement from minimal bar diameter 9.5.3(1)
Based on minimal bar diameter requirement which is defined in clause 9.5.3(1). It can be transformed to formula
ϕt,min,col = max(dsc,min; xdsc ∙ ϕl)
where
dsc,min | Minimal diameter of transverse reinforcement according to clause 9.5.3(1) |
xdsc | Multiplication of the maximum diameter of the longitudinal bars |
ϕl | Maximal diameter of the longitudinal bars |
Final minimal area is calculated by
Aswc,ϕ,min = π ∙ ϕt,min,col2 / 4 ∙ ns
where
ϕt,min,col | Minimal diameter of transverse reinforcement according to clause 9.5.3(1) |
ns | Number of cuts in one shear link |
Minimal area of shear reinforcement from maximum longitudinal spacing 9.5.3(3)
Based on minimal allowed longitudinal spacing of transversal reinforcement which is defined in clause 9.5.3(3). It can be transformed to formula
scl,tmax = min(Coeffcl,max,A ∙ ϕl; bmin; Coeffcl,max,B)
where
Final minimal area is calculated by
Aswc,long,min = 1 / sscl,tmax ∙ π ∙ ϕ2 / 4 ∙ ns
where
sscl,tmax | Maximum longitudinal spacing of transversal reinforcement according to clause 9.5.3(3) |
ϕ | Diameter of shear reinforcement |
ns | Number of cuts in one shear link |
The values of coefficients or calculation procedures used in all detailing provisions can be modified by National annexes, for details see "National annexes theoretical background".